Technical Calculus Worksheet–Optimization
Set up, but do not solve the following problems. In other words, find a function of one
variable with an appropriate domain that you would find the maximum or minimum of in
order to solve the problem.
1. No group did this question.
2. A rectangular storage container with an open top is to have a volume of 10 m3 . The
length of its base is twice the width. Material for the base costs $10 per square meter.
Material for the sides costs $6 per square meter. Find the cost of materials for the
cheapest such container.
Solution: Slit along the height of the box and flatten it out, we have a diagram that
looks like:
h
h
2w
h
w
h
The total cost, C = cost of base + cost of 4 sides, i.e.
C = 10(area of base) + 6(area of 4 sides)
= 10(2w2 ) + 6(wh + 2wh + 2wh + wh)
= 20w2 + 36wh ( Pri. Eqn.)
Since the volume is 10, we get
10 = (2w)(w)(h) = 2w2 h ( Sec. Eqn. )
Solving for h yield h =
5
w2
and substituting this into the primary equation, we get
5
C = 20w + 36w
w2
2
= 20w2 +
180
w
Now we want to minimize C as w varies between 0 and ∞.
3. A new running track will be built. It is to be the perimeter of a region obtained by
putting two semicircles on the ends of a rectangle. If the track is to be 440 yards long,
determine the necessary dimensions to build the track in order to maximize the area
the track enclose
Solution: The diagram should look something like:
$
'
y
&
%
x
It can be seen that the radius of the semi-circle is y/2, hence area of the two semi-circles
is π(y/2)2 = 41 πy 2 .
We want to maximize the area enclosed, so the enclosed area, A = 14 πy 2 + xy (primary
equation.)
The constraint is, the track is only 440 yards, so
y
y
440 = x + π
+x+π
2
2
secondary eqn: 440 = 2x+πy ⇒ x = 440−πy
. Which when substituted into the primary
2
equation yields:
1
440 − πy
1
= 220y − πy 2
A = πy 2 + y
4
2
4
0 ≤ y ≤ R where R is some number about 220 (otherwise, y will be too big and there
will not be a feasible x value.)
The critical number of that function is y = 880
.
π
4. [No group do this problem.]
5. A feed-lot farmer has 1000 feet of fencing with which to enclose two adjacent rectangular corrals. What dimensions should be used so that the enclosed area will be a
maximum?
2
Solution: The diagram should look something like:
w
L
w
L
L
w
w
The enclosed area, A is given by
A = 2wL (Pri. Eqn.).
Since we only have 1000 feet of fence
1000 = 3L + 4w
Solving for L yield L =
get
1000−4w
3
( Sec. Eqn. )
and substituting this into the Primary Equation, we’ll
1000 − 4w
A = 2w
3
Now we really want to maximize A as w varies between 0 and 250. (You get 250
because we want L which is 13 (100 − 4w) to be positive.)
6. A wire 2 meters long is cut into two pieces. One piece is bent into a square for a stained
glass frame while the other piece is bent into a circle for a TV antenna. Where should
the wire be cut to maximize the total area?
Let the first wire be of length x and the second wire be of length y.
the first wire made into a square would have sides x4 while the second wire would have
y
radius 2π
.
2
2
y
(primary equation).
The total area, A = x4 + π 2π
Since the wire is only 2 meters long, Our secondary equation is: x + y = 2.
substituting y = 2 − x into the Primary Equation, we’ll get
1 2
1
A =
x +
(2 − x)2
16
4π
dA
1
1
⇒
=
x−
(2 − x)
dx
8
2π
2
is a crit. number
⇒x =
1 + π4
3
7. [No group did this problem]
8. A Norman window has the shape of a rectangle surmounted by a semicircle. Thus,
the diameter of the semicircle is equal to the width of the the rectangle.) If the
perimeter of the window is 30 ft, find the dimensions of the window so that the greatest
possible amount of light is admitted. A Norman window has the shape of a rectangle
surmounted by a semicircle. Thus, the diameter of the semicircle is equal to the width
of the the rectangle.)
'$
y
x
Solution:
It can be seen that the radius of the semi-circle is x/2, hence area of the semi-circle is
1
π(x/2)2 = 81 πx2 .
2
The total area, A = 18 πx2 + xy (Pri. eqn.)
Since the perimeter has to be 30 ft, from this constraint, we get
x
2
30 = y + x + y + π
( Sec. Eqn. )
Solving for y, we get
πx
2
2x + πx
⇒ y = 15 −
4
2y = 30 − x −
Substituting this into the primary equation we get
1
2x + πx
A = πx2 + x 15 −
8
4
We wish to maximize A as x
varies between 0 and some number slightly less than
4
it comes from requiring y = 15 − 2x+πx
15 (the actual number is 15
to be
4
2+π
positive.)
4